Why is the derivative 0 if the policy is deterministic?
First, a small clarification; they do not say that "the gradient of a deterministic policy is 0" (I'm not sure what, if anything, "the gradient of a policy") would mean. They are talking about how "this gradient" (referring to the gradient on the slide visible in the video) is 0 in the case where the policy ($\pi$) is deterministic.

Now, here is the gradient that they are discussing in the video:

$$\nabla_{\theta} J(\theta) \approx \frac{1}{N} \sum_{i=1}^N \left( \sum_{t=1}^T \nabla_{\theta} \log \pi_{\theta} (\mathbf{a}_{i, t} \vert \mathbf{s}_{i, t}) \right) \left( \sum_{t = 1}^T r(\mathbf{s}_{i,t}, \mathbf{a}_{i, t}) \right)$$

In this equation, $\pi_{\theta} (\mathbf{a}_{i, t} \vert \mathbf{s}_{i, t})$ denotes the probability of our policy $\pi_{\theta}$ selecting the actions $\mathbf{a}_{i, t}$ that it actually ended up selecting in practice, given the states $\mathbf{s}_{i, t}$ that it encountered during the episode that we're looking at.

In the case of a deterministic policy $\pi_{\theta}$, we know for sure that the probability of it selecting the actions that it did select must be $1$ (and the probability of it selecting any other actions would be $0$, but such a term does not show up in the equation). So, we have $\pi_{\theta} (\mathbf{a}_{i, t} \vert \mathbf{s}_{i, t}) = 1$ for every instance of that term in the above equation. Because $\log 1 = 0$, this leads to:

\begin{aligned}
\nabla_{\theta} J(\theta) &amp;\approx \frac{1}{N} \sum_{i=1}^N \left( \sum_{t=1}^T \nabla_{\theta} \log \pi_{\theta} (\mathbf{a}_{i, t} \vert \mathbf{s}_{i, t}) \right) \left( \sum_{t = 1}^T r(\mathbf{s}_{i,t}, \mathbf{a}_{i, t}) \right) \\
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&amp;= \frac{1}{N} \sum_{i=1}^N \left( \sum_{t=1}^T \nabla_{\theta} \log 1 \right) \left( \sum_{t = 1}^T r(\mathbf{s}_{i,t}, \mathbf{a}_{i, t}) \right) \\
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&amp;= \frac{1}{N} \sum_{i=1}^N \left( \sum_{t=1}^T \nabla_{\theta} 0 \right) \left( \sum_{t = 1}^T r(\mathbf{s}_{i,t}, \mathbf{a}_{i, t}) \right) \\
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&amp;= \frac{1}{N} \sum_{i=1}^N 0 \left( \sum_{t = 1}^T r(\mathbf{s}_{i,t}, \mathbf{a}_{i, t}) \right) \\
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&amp;= 0 \\
\end{aligned}

(i.e. you end up with a sum of terms that are all multiplied by $0$).